Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
double solve(double x, double y, double ax, double ay) {
  return y / (y - ay) * (ax - x) + x;
}
int n2;
int a[N << 1], sum[N << 1];
int bin(int x) {
  int l = 0, r = n2;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (a[mid] > x)
      r = mid - 1;
    else
      l = mid;
  }
  return l;
}
int main() {
  int sy, ax, bx, ll, rr, dl, dr, ans, T, n, x, y;
  double l, r, res;
  scanf("%d%d%d", &sy, &ax, &bx);
  scanf("%d", &n);
  n2 = n * 2;
  for (int i = 1; i <= n2; i += 2) scanf("%d%d", &a[i], &a[i + 1]);
  sum[0] = 0;
  for (int i = 1; i <= n2; i++) {
    if (i & 1)
      sum[i] = sum[i - 1];
    else
      sum[i] = sum[i - 1] + a[i] - a[i - 1];
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &x, &y);
    l = solve((double)x, (double)y, (double)ax, (double)sy);
    r = solve((double)x, (double)y, (double)bx, (double)sy);
    ll = int(l);
    rr = int(r);
    dl = bin(ll);
    dr = bin(rr);
    ans = sum[dr] - sum[dl];
    res = 0.0;
    if (dl & 1) {
      res -= l - (double)ll;
      ans -= ll - a[dl];
    }
    if (dr & 1) {
      res += r - (double)rr;
      ans += rr - a[dr];
    }
    res += (double)ans;
    res = res * ((double)bx - (double)(ax)) / (r - l);
    printf("%.10f\n", res);
  }
}